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Section 10.7: Taylor and Maclaurin Series
Taylor and Maclaurin series are power series representations of functions. Let
f(x) =∞∑n=0
cn(x− a)n = c0 + c1(x− a) + c2(x− a)2 + c3(x− a)3 + c4(x− a)4 + · · · .
Then f(a) = c0.
Taking the derivative gives
f ′(x) = c1 + 2c2(x− a) + 3c3(x− a)2 + 4c4(x− a)3 + · · · .
Then f ′(a) = c1.
Similarly,f ′′(x) = 2c2 + 3 · 2c3(x− a) + 4 · 3c4(x− a)2 + · · · .
Then f ′′(a) = 2c2 and so c2 =f ′′(a)
2.
Likewise,f ′′′(x) = 3 · 2c3 + 4 · 3 · 2(x− a) + · · · .
Then f ′′′(a) = 3!c3 and so c3 =f ′′′(a)
3!.
Continuing in this fashion, we find that in general
cn =f (n)(a)
n!.
Definition: The Taylor series for f(x) centered at x = a is
f(x) =∞∑n=0
f (n)(a)
n!(x− a)n,
where f (n)(a) is the nth derivative of f(x) at x = a.
Example: Find the Taylor series for f(x) = e3x centered at x = 2. What is the associatedradius of convergence?
The higher-order derivatives of f(x) are
f(x) = e3x f(2) = e6
f ′(x) = 3e3x f ′(2) = 3e6
f ′′(x) = 9e3x f ′′(2) = 9e6
f ′′′(x) = 27e3x f ′′′(2) = 27e6
......
f (n)(x) = 3ne3x f (n)(2) = 3ne6.
So the Taylor series is
e3x =∞∑n=0
3ne6
n!(x− 2)n.
Using the Ratio Test,
limn→∞
∣∣∣∣3n+1e6(x− 2)n+1
(n+ 1)!· n!
3ne6(x− 2)n
∣∣∣∣ = limn→∞
∣∣∣∣3(x− 2)
n+ 1
∣∣∣∣ = 0 < 1.
The radius of convergence is R =∞.
Example: Find the Taylor series for f(x) =1
xcentered at x = 3. What is the associated
radius of convergence?
The higher-order derivatives of f(x) are
f(x) =1
xf(3) =
1
3
f ′(x) = − 1
x2f ′(3) = −1
9
f ′′(x) =2
x3f ′′(3) =
2
27
f ′′′(x) = − 6
x4f ′′′(3) = − 6
81...
...
f (n)(x) =(−1)nn!
xn+1f (n)(3) =
(−1)nn!
3n+1.
So the Taylor series is1
x=∞∑n=0
(−1)n
3n+1(x− 3)n.
Using the Ratio Test,
limn→∞
∣∣∣∣(−1)n+1(x− 3)n+1
3n+2· 3n+1
(−1)n(x− 3)n
∣∣∣∣ = limn→∞
∣∣∣∣−(x− 3)
3
∣∣∣∣ =|x− 3|
3< 1.
The radius of convergence is R = 3.
Definition: The Maclaurin series for f(x) is the Taylor series centered at x = 0. That is,
f(x) =∞∑n=0
f (n)(0)
n!xn,
where f (n)(0) is the nth derivative of f(x) at x = 0.
Example: Find the Maclaurin series for f(x) = ex. What is the associated radius of conver-gence?
For n ≥ 0, the nth derivative of f(x) is
f (n)(x) = f(x) = ex.
Then f (n)(0) = 1 and the Maclaurin series is
ex =∞∑n=0
xn
n!.
Using the Ratio Test,
limn→∞
∣∣∣∣ xn+1
(n+ 1)!· n!
xn
∣∣∣∣ = limn→∞
∣∣∣∣ x
n+ 1
∣∣∣∣ = 0 < 1.
The radius of convergence is R =∞.
Example: Find the Maclaurin series for f(x) = sinx. What is the associated radius ofconvergence?
The higher-order derivatives of f(x) = sin x are
f(x) = sinx f(0) = 0f ′(x) = cosx f ′(0) = 1f ′′(x) = − sinx f ′′(0) = 0f ′′′(x) = − cosx f ′′′(0) = −1f (4)(x) = sinx f (4)(0) = 0
......
The Maclaurin series for sinx is
sinx = x− x3
3!+x5
5!− x7
7!+− · · · =
∞∑n=0
(−1)nx2n+1
(2n+ 1)!.
Using the Ratio Test,
limn→∞
∣∣∣∣(−1)n+1x2n+3
(2n+ 3)!· (2n+ 1)!
(−1)nx2n+1
∣∣∣∣ = limn→∞
∣∣∣∣ −x2
(2n+ 3)(2n+ 2)
∣∣∣∣ = 0 < 1.
The radius of convergence is R =∞.
Example: Find the Maclaurin series for f(x) = cosx. What is the associated radius ofconvergence?
The higher-order derivatives of f(x) = cos x are
f(x) = cosx f(0) = 1f ′(x) = − sinx f ′(0) = 0f ′′(x) = − cosx f ′′(0) = −1f ′′′(x) = sinx f ′′′(0) = 0f (4)(x) = cosx f (4)(0) = 1
......
The Maclaurin series for cosx is
cosx = 1− x2
2!+x4
4!− x6
6!+− · · · =
∞∑n=0
(−1)nx2n
(2n)!.
Using the Ratio Test,
limn→∞
∣∣∣∣(−1)n+1x2n+2
(2n+ 2)!· (2n)!
(−1)nx2n
∣∣∣∣ = limn→∞
∣∣∣∣ −x2
(2n+ 2)(2n+ 1)
∣∣∣∣ = 0 < 1.
The radius of convergence is R =∞.
Example: Find the Maclaurin series for f(x) = ex3. What is the associated radius of conver-
gence?
The Maclaurin series for eu is
eu =∞∑n=0
un
n!.
Setting u = x3, the Maclaurin series for ex3
is
ex3
=∞∑n=0
(x3)n
n!=∞∑n=0
x3n
n!.
The radius of convergence is R =∞.
Example: Find the Maclaurin series for f(x) = x cos(x3). What is the associated radius ofconvergence?
The Maclaurin series for cosu is
cosu =∞∑n=0
(−1)nu2n
(2n)!.
Setting u = x3, the Maclaurin series for x cos(x3) is
x cos(x3) = x∞∑n=0
(−1)n(x3)2n
(2n)!=∞∑n=0
(−1)nx6n+1
(2n)!.
The radius of convergence is R =∞.
Example: Find the Maclaurin series for f(x) = x2 sin(x
2
). What is the associated radius of
convergence?
The Maclaurin series for sinu is
sinu =∞∑n=0
(−1)nu2n+1
(2n+ 1)!.
Setting u =x
2, the Maclaurin series for x2 sin
(x2
)is
x sin(x
2
)= x
∞∑n=0
(−1)n
(2n+ 1)!
(x2
)2n+1
=∞∑n=0
(−1)nx2n+2
22n+1(2n+ 1)!.
The radius of convergence is R =∞.
Example: Consider the definite integral
∫ 1
0
cos(x2)dx.
(a) Evaluate the integral as a series.
The Maclaurin series for cos(x2) is
cos(x2) =∞∑n=0
(−1)n(x2)2n
(2n)!=∞∑n=0
(−1)nx4n
(2n)!.
Then ∫ 1
0
cos(x2)dx =∞∑n=0
(−1)nx4n+1
(4n+ 1)(2n)!
∣∣∣∣∣1
0
=∞∑n=0
(−1)n
(4n+ 1)(2n)!.
(b) Use the sum of the first three terms to approximate the integral. How accurate is thisapproximation?
The second partial sum (sum of the first three terms) is∫ 2
0
cos(x2)dx ≈ S2 = 2− 1
10+
1
216.
By the Remainder Estimate for Alternating Series,
|R2| ≤ a3 =1
13(6)!=
1
9360≈ 0.00011.
Example: Approximate
∫ 1
0
e−x2
dx with error less than 0.001.
The Maclaurin series for e−x2
is
e−x2
=∞∑n=0
(−x2)n
n!=∞∑n=0
(−1)nx2n
n!.
Then ∫ 1
0
e−x2
dx =∞∑n=0
(−1)nx2n+1
(2n+ 1)(n)!
∣∣∣∣∣1
0
=∞∑n=0
(−1)n
(2n+ 1)(n)!.
By the Remainder Estimate for Alternating Series,
|Rn| ≤ an+1 =1
(2n+ 3)(n+ 1)!.
Let
1
(2n+ 3)(n+ 1)!<
1
1000
(2n+ 3)(n+ 1)! > 1000.
If n = 4, then (2n+ 3)(n+ 1)! = (11)5! = 1320 > 1000. Then∫ 1
0
e−x2
dx ≈ S4 = 1− 1
3+
1
10− 1
42+
1
216≈ 0.7475.
The true value is approximately 0.7469.
Example: Evaluate limx→0sin(x3)− x3
x9.
The Maclaurin series for sin(x3) is
∞∑n=0
(−1)n(x3)2n+1
(2n+ 1)!=
∞∑n=0
(−1)nx6n+3
(2n+ 1)!
= x3 − x9
3!+x15
5!−+ · · · .
Then
limx→0
sin(x3)− x3
x9= lim
x→0
−x9
3!+ x15
5!−+ · · ·
x9
= limx→0
(− 1
3!+x6
5!−+ · · ·
)= −1
6.
Example: Find the sum of the given series.
(a)∞∑n=0
(−1)n3nx2n
n!
The series can be rewritten as
∞∑n=0
(−3x2)n
n!= e−3x
2
.
(b)∞∑n=0
(−1)nπ2n+1
32n+1(2n+ 1)!
The series can be rewritten as
∞∑n=0
(−1)n
(2n+ 1)!
(π3
)2n+1
= sin(π
3
)=
√3
2.
(c)∞∑n=1
(−4)nx2n
n!
The series can be rewritten as
∞∑n=1
(−4x2)n
n!=∞∑n=0
(−4x2)n
n!− 1 = e−4x
2 − 1.
